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Full text of "Text Book Of Mechanical Engineering"

Position of Neutral Axis.

in compression and BD in tension, The stresses will be zero
at B and increase towards c and D as shewn, forming a couple,
(j or F) >c G H, to resist bending, where j — F. Consider two
small areas at and a^ and let p = radius of curvature at neutral
line. Then:

Length before bending               = 2 T p

Length of ring at after bending   =2x1
Length of ring ac after bending   = 2 T (p -j
,*.  Strain on fibre at— 2 TT (p+-)'t) — 2 TT p   = 2 T yt
and Strain on fibre czc~ 2 T p - 2 rr(p-jc)   = 2 ?ryc

and/=^...
1°

But A =J~ generally   . •. 2 Try =*J—

•tL

Total stress on a small area = /j = ±i^L"

/°

E Tt ^t

Total stress on area BIDI = sum of —— for all portions of BL r>j

f>

and Total stress on area BJ Cj_ = sum of   " °  c for all portions of \ Cj

But these are the forces F and j,

and as — is a constant,

P

SjK ^t=2j/c«c........................  (2)

or      Moment of tension area = moment of compression area.

But the centre of gravity of a lamina or centre of figure of area
is such that the. moments on either side are equal. Therefore tht
neutral axis of any bar passes through the centre ojjigure (or centra id)
of Us cross section.

Moment of Resistance.—^Again, in Fig. 383.

Eya

Moment of stress on a =

Moment of all stresses on a section

x y

P         J     P

But   2<3j/2 is the moment of inertia (2nd moment) of the
section = 1

"F T
.*.   Moment of resistance = -— (in terms of p)...... (3)

(Seep. 107.8.)